#### provide solution for RD Sharma maths class 12 chapter determinant exercise 5.5 question 1

$x = k, y = k, z = k, where \: k \in R$

Hint: The homogeneous system of linear equations has non-trivial solutions when the determinant is zero.

Given:

$x + y - 2z =0$

$2x + y - 3z = 0$

$5x + 4y - 9z = 0$

Solution:

The above linear equations can be represented in matrix form i.e.

$A = \begin{bmatrix} 1& 1& -2\\ 2& 1& -3\\ 5& 4& -9 \end{bmatrix}$, $X= \begin{bmatrix} x\\ y\\ z\end{bmatrix}$,$B= \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$

Now,        $A = \begin{vmatrix} 1 & 1&-2 \\ 2& 1& -3\\ 5& 4& -9 \end{vmatrix}$

$= 1(-9 + 12) -1(-18 + 15) - 2(8 -5)$

$=3 + 3 - 6$

$=0$

So, the given system has non-trivial solutions.

Consider first two equations and put z=k.

$x+ y - 2z =0$

$2x + y -3z = 0$

Solving these equations by Crammer’s rule:

$x = \frac{D_{1}}{D}=\frac{\begin{vmatrix} 2k&1 \\ 3k& 1 \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k$

$y = \frac{D_{2}}{D}=\frac{\begin{vmatrix} 1&2k \\ 2& 3k \end{vmatrix}}{\begin{vmatrix} 1 & 1\\ 2& 1 \end{vmatrix}} = \frac{-k}{-1}=k$

Thus, $x = k, y = k, z = k$ satisfy the third equation.

Hence,$x = k, y = k, z = k$  where $k \in R$